# Full text of "Treatise On Analysis Vol-Ii"

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```3 THE MODULUS FUNCTION ON A GROUP       261

n

such that £ x? £ 1. The same method that was used for the calculation of
Hn(Sn) gives

n- 1
"-1   r.-,i^U   +Uo+  V    V2 ^ i        i2

Now Bn(/l) is the set of all (xl9 ..., xn^) e R"""1 such that £ x? <; 1 - A2,

and hence is obtained from Rn-i by the homethety with ratio (1— A2)1/2,
and therefore

Making the change of variable A = sin 0, we get
(14.3.11.1)     Vfl = Vn_1 f/2 cosn0J0=2Vn_1 \

J-n/2                               JO

cos* Od8.

Put cn = r   cos"0 d9, and apply the formula of integration by parts (8.7.5);
then if n > 2 we obtain

fn/2

• = (»-l)       c
Jo

cos"""2 9 sin2 OdO=(n- l)(cn^2 - crt),

so that ncn = (n — l)cn^2 • Since c0 = ^TT and q = 1, we have finally

1 • 3 • 5 • • - (2n - 1)   TT                  _ 2 • 4 - 6 • • • (2n - 2)

C2n~     2'4^6'"2n     %2'       C2r>~1 "" 1 • 3 • 5 • •• (2n - 1)"

Since Vt = 2 and V2 = it we obtain from (14.3.11.1)

(14.3.11.2)

7C"

n-l

In terms of the gamma function,* these formulas can be written in the form
(14.3.11.3)                          Vn = nn/2/r(\$n + 1).

* See for example my book, Calcul infinitesimal (Hermann, Paris, 1968). e R""1 such that
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